Here $\mu$ denotes the Lebesgue measure on $[0,1]$. WLOG, assume $g=0$. Then we are given with $$\int\limits_A f\ d\mu=0\ \forall A\subset[0,1]\text{ with }\mu(A)=1/3$$
First I assume that $f\ge0$. Then I divide $[0,1/3]$ into three intervals of measure $1/3$ say $[0,1/3],[1/3,2/3]$ and $[2/3,1]$. Let's first restrict to $I_1=[0,1/3]$. We have that $\int\limits_{I_1}f\ d\mu=0$. Since, $f\ge0$, we have that $\int\limits_Af\ d\mu=0$ for all $A\subseteq I_1$.
Therefore, $\int\limits_A f\chi_{I_1}\ d\mu=0$ for all measurable subsets of $I_1$. Hence, $f=0$ a.e. on $I_1$. Similarly, $f=0$ a.e. on other two intervals as well. Hence, $f=0$ a.e.
Now for general real valued measurable $f$, this process won't work. For that I define $$\mathcal{M}=\left\{A\subseteq[0,1]:\ A\text{ is measurable, }\int\limits_Af\ d\mu=0\right\}$$
I can prove that it is a monotone class and contains all measurable subsets of measure $1/3$. I want to apply Monotone Class Theorem here, for that I need to show at least all the intervals should be in $\mathcal{M}$. But I am stuck here.
Can anyone help me to finish the proof? Or, should I take a different approach for the general case? Thanks for your help in advance.
